bài 1 tính A=(1+1/2).(1+1/3).(1+1/4)…(1+1/99) B= 3/4.8/9.15/16…899/900 03/07/2021 Bởi Josephine bài 1 tính A=(1+1/2).(1+1/3).(1+1/4)…(1+1/99) B= 3/4.8/9.15/16…899/900
Ta có: A=(1+$\frac{1}{2}$)(1+$\frac{1}{3}$)(1+$\frac{1}{4}$)……(1+$\frac{1}{99}$) =($\frac{2}{2}$+$\frac{1}{2}$)($\frac{3}{3}$+$\frac{1}{3}$)($\frac{4}{4}$+$\frac{1}{4}$)+……+( $\frac{99}{99}$+$\frac{1}{99}$) = $\frac{3}{2}$.$\frac{4}{3}$.$\frac{5}{4}$……….$\frac{100}{99}$ =$\frac{3.4.5…..100}{2.3.4…..99}$ = $\frac{100}{2}$=50 B=$\frac{3}{4}$.$\frac{8}{9}$.$\frac{15}{16}$…….$\frac{899}{900}$ =$\frac{1.3}{2.2}$.$\frac{2.4}{3.3}$.$\frac{3.5}{4.4}$…..$\frac{29.31}{30.30}$ = $\frac{1.2.3…..29}{2.3.4…..30}$.$\frac{3.4.5…..31}{2.3.4…..30}$ = $\frac{1}{30}$.$\frac{31}{2}$=$\frac{31}{60}$ Chúc học tốt!!! Bình luận
A=(1+$\frac{1}{2}$).(1+$\frac{1}{3}$).(1+$\frac{1}{4}$)+…+(1+ $\frac{1}{99}$)=($\frac{2}{2}$ +$\frac{1}{2}$).($\frac{3}{3}$+$\frac{1}{3}$).($\frac{4}{4}$+$\frac{1}{4}$)+…+($\frac{99}{99}$+$\frac{1}{99}$)= $\frac{3}{2}$.$\frac{4}{3}$.$\frac{5}{4}$+…+$\frac{100}{99}$=$\frac{100}{2}$ =50B=$\frac{3}{4}$.$\frac{8}{9}$.$\frac{15}{16}$….$\frac{899}{900}$ =$\frac{1.3}{2^2}$.$\frac{2.4}{3^3}$.$\frac{3.5}{4^2}$….$\frac{29.31}{30^2}$ =$\frac{(1.2.3…29)(3.4.5.31)}{(2.3.4….30)(2.3.4….30)}$=$\frac{1.2.3…29}{2.3.4….30}$.$\frac{3.4.5…31}{2.3.4….30}$=$\frac{1}{30}$.$\frac{31}{2}$ =$\frac{31}{60}$. Bình luận
Ta có:
A=(1+$\frac{1}{2}$)(1+$\frac{1}{3}$)(1+$\frac{1}{4}$)……(1+$\frac{1}{99}$)
=($\frac{2}{2}$+$\frac{1}{2}$)($\frac{3}{3}$+$\frac{1}{3}$)($\frac{4}{4}$+$\frac{1}{4}$)+……+( $\frac{99}{99}$+$\frac{1}{99}$)
= $\frac{3}{2}$.$\frac{4}{3}$.$\frac{5}{4}$……….$\frac{100}{99}$
=$\frac{3.4.5…..100}{2.3.4…..99}$
= $\frac{100}{2}$=50
B=$\frac{3}{4}$.$\frac{8}{9}$.$\frac{15}{16}$…….$\frac{899}{900}$
=$\frac{1.3}{2.2}$.$\frac{2.4}{3.3}$.$\frac{3.5}{4.4}$…..$\frac{29.31}{30.30}$
= $\frac{1.2.3…..29}{2.3.4…..30}$.$\frac{3.4.5…..31}{2.3.4…..30}$
= $\frac{1}{30}$.$\frac{31}{2}$=$\frac{31}{60}$
Chúc học tốt!!!
A=(1+$\frac{1}{2}$).(1+$\frac{1}{3}$).(1+$\frac{1}{4}$)+…+(1+ $\frac{1}{99}$)
=($\frac{2}{2}$ +$\frac{1}{2}$).($\frac{3}{3}$+$\frac{1}{3}$).($\frac{4}{4}$+$\frac{1}{4}$)+…+($\frac{99}{99}$+$\frac{1}{99}$)
= $\frac{3}{2}$.$\frac{4}{3}$.$\frac{5}{4}$+…+$\frac{100}{99}$
=$\frac{100}{2}$
=50
B=$\frac{3}{4}$.$\frac{8}{9}$.$\frac{15}{16}$….$\frac{899}{900}$
=$\frac{1.3}{2^2}$.$\frac{2.4}{3^3}$.$\frac{3.5}{4^2}$….$\frac{29.31}{30^2}$
=$\frac{(1.2.3…29)(3.4.5.31)}{(2.3.4….30)(2.3.4….30)}$
=$\frac{1.2.3…29}{2.3.4….30}$.$\frac{3.4.5…31}{2.3.4….30}$
=$\frac{1}{30}$.$\frac{31}{2}$
=$\frac{31}{60}$.